Solve ∫ (x^3-x)(3x^2-1) wrt x

$$\int ( x ^ { 3 } - x ) ( 3 x ^ { 2 } - 1 ) d x$$

Evaluate

$$\frac{x^{6}}{2}-x^{4}+\frac{x^{2}}{2}+С$$

Solution Steps

Use the distributive property to multiply x^{3}-x by 3x^{2}-1 and combine like terms.

$$\int 3x^{5}-4x^{3}+x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{5}\mathrm{d}x+\int -4x^{3}\mathrm{d}x+\int x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{5}\mathrm{d}x-4\int x^{3}\mathrm{d}x+\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply 3 times \frac{x^{6}}{6}.

$$\frac{x^{6}}{2}-4\int x^{3}\mathrm{d}x+\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -4 times \frac{x^{4}}{4}.

$$\frac{x^{6}}{2}-x^{4}+\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}.

$$\frac{x^{6}}{2}-x^{4}+\frac{x^{2}}{2}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{x^{6}}{2}-x^{4}+\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

$$3x^{5}-4x^{3}+x$$